Question 368750: In the department of education at UR university, student records suggest that the population of students spends an average of 5.5 hrs per week playing organized sports. The population's standard deviation is 2.2 hrs per week. Based on a sample of 121 students, healthy lifestyles Inc. (HLI) would like to apply the central limit theorem to make various estimates.
Compute the standard error of the sample mean
what is the chance HLI will find a sample mean between 5 and 6 hrs?
calculate the probability that the sample mean will be between 5.3 and 5.7 hrs
how strange would it be to obtain a sample mean greater than 6.5 hrs?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! In the department of education at UR university, student records suggest that the population of students spends an average of 5.5 hrs per week playing organized sports. The population's standard deviation is 2.2 hrs per week.
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Based on a sample of 121 students, healthy lifestyles Inc.
(HLI) would like to apply the central limit theorem to make various estimates.
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Compute the standard error of the sample mean?
Ans: 2.2/sqrt(121) = 0.2
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What is the chance HLI will find a sample mean between 5 and 6 hrs?
t(5) = (5-5.5)/0.2 = -2.5
t(6) = (6-5.5)/0.2 = +2.5
P(5< x-bar <6) = P(-2.5< t < 2.5 when df=120) = 0.9862
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Calculate the probability that the sample mean will be between 5.3 and 5.7 hrs
how strange would it be to obtain a sample mean greater than 6.5 hrs?
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Find the t-values; then find the probabilities.
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Cheers,
Stan H.
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